Prove that tan theta+tan 2theta+tan 3theta=tan theta*tan 2theta*tan 3theta
I think you have it wrong:
tan 3θ = tan(θ+2θ)
tan 3θ = (tanθ+tan2θ)/(1-tanθtan2θ)
tan3θ(1-tanθtan2θ) = tanθ+tan2θ
tan3θ - tanθtan2θtan3θ = tanθ+tan2θ
tan3θ - tan2θ - tanθ = tanθtan2θtan3θ
To prove the given trigonometric identity:
tan(theta) + tan(2theta) + tan(3theta) = tan(theta) * tan(2theta) * tan(3theta),
we will start by expressing both sides of the equation in terms of sine and cosine.
First, let's express the left side of the equation in terms of sine and cosine:
tan(theta) + tan(2theta) + tan(3theta)
Using the identity tan(x) = sin(x)/cos(x), we can rewrite each term in terms of sine and cosine:
sin(theta)/cos(theta) + sin(2theta)/cos(2theta) + sin(3theta)/cos(3theta)
Next, we'll find a common denominator for the three fractions:
[(sin(theta)*cos(2theta)*cos(3theta)) + (sin(2theta)*cos(theta)*cos(3theta)) + (sin(3theta)*cos(theta)*cos(2theta))] / (cos(theta)*cos(2theta)*cos(3theta))
Now, we can simplify the numerator:
[(sin(theta)*cos(2theta)*cos(3theta)) + (sin(2theta)*cos(theta)*cos(3theta)) + (sin(3theta)*cos(theta)*cos(2theta))]
By applying the product-to-sum identities (sin(A + B) = sin(A)cos(B) + cos(A)sin(B) and cos(A + B) = cos(A)cos(B) - sin(A)sin(B)), we can rewrite the numerator as:
[sin(theta)*(cos(2theta)*cos(3theta))] + [sin(2theta)*(cos(theta)*cos(3theta))] + [sin(3theta)*(cos(theta)*cos(2theta))]
Expanding further by applying the double angle identities (cos(2theta) = cos^2(theta) - sin^2(theta) and sin(2theta) = 2sin(theta)cos(theta)), we get:
[sin(theta)*((cos^2(theta) - sin^2(theta))*cos(3theta))] + [2sin(theta)*cos(theta)*(cos(theta)*cos(3theta))] + [sin(3theta)*(cos(theta)*cos^2(theta) - sin^2(theta)*cos(theta))]
Further simplifying each term, we get:
[sin(theta)*(cos^2(theta)*cos(3theta) - sin^2(theta)*cos(3theta))] + [2sin(theta)*cos^2(theta)*cos(3theta)] + [sin(3theta)*(cos(theta)*cos^2(theta) - sin^2(theta)*cos(theta))]
Using the formula sin^2(theta) = 1 - cos^2(theta), we can rewrite this as:
[sin(theta)*cos^2(theta)*cos(3theta) - sin^2(theta)*cos^3(3theta)] + [2sin(theta)*cos^2(theta)*cos(3theta)] + [sin(3theta)*(cos(theta)*cos^2(theta) - (1 - cos^2(theta))*cos(theta))]
Simplifying, we have:
[sin(theta)*cos^2(theta)*cos(3theta) - sin^2(theta)*cos^3(3theta)] + [2sin(theta)*cos^2(theta)*cos(3theta)] + [sin(3theta)*(2cos^3(theta) - cos(theta))]
Now, let's consider the right side of the equation:
tan(theta) * tan(2theta) * tan(3theta)
Using the identity tan(x) = sin(x)/cos(x), we can express this as:
(sin(theta)/cos(theta)) * (sin(2theta)/cos(2theta)) * (sin(3theta)/cos(3theta))
Now, let's simplify this expression by multiplying the numerators and denominators:
[(sin(theta)*sin(2theta)*sin(3theta)) / (cos(theta)*cos(2theta)*cos(3theta))]
Comparing the simplified expression of the right side to the numerator we obtained earlier while simplifying the left side, we can see that they are identical. Therefore:
tan(theta) + tan(2theta) + tan(3theta) = tan(theta) * tan(2theta) * tan(3theta),
and the given trigonometric identity is proven.